13(z+3)-5(z-3)=4(z-2)+3(z-2)

Solution for 13(z+3)-5(z-3)=4(z-2)+3(z-2) equation:

13(z+3)-5(z-3)=4(z-2)+3(z-2)

We move all terms to the left:

13(z+3)-5(z-3)-(4(z-2)+3(z-2))=0

We multiply parentheses

13z-5z-(4(z-2)+3(z-2))+39+15=0


We calculate terms in parentheses: -(4(z-2)+3(z-2)), so:

4(z-2)+3(z-2)

We multiply parentheses

4z+3z-8-6

We add all the numbers together, and all the variables

7z-14

Back to the equation:

-(7z-14)


We add all the numbers together, and all the variables

8z-(7z-14)+54=0

We get rid of parentheses

8z-7z+14+54=0

We add all the numbers together, and all the variables

z+68=0

We move all terms containing z to the left, all other terms to the right

z=-68